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A 100muF capacitor in series with a 40Omega resistance is connected to 110 V, 60 Hz supply . a. What is the maximum current in the circuit ? b. What isthe time lag between the current maximum and the voltage maximum ? |
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Answer» Solution :Data supplied`C= 100 mu F, R = 40 Omega , E_(rms) = 110 V , v = 69Hz` ` E_0 = E_(rms)XX sqrt(2) = 110 xx 1.414 = 155.6 V` a.` i_m (E_0)/(Z) = (E_0)/(sqrt(R^2 + ( (1)/(omega C) )^2)) = (155.6 )/(sqrt(40^2 + ( (1)/(10^(-4) xx 2 xx 3.14 xx 69) )^2) ) = 3.242 A` B. ` tan phi = (1)/(omega CR)` ` tan phi = (1)/(10^(-4) xx 2 xx 3.14 xx 69 xx 40)= 0.6635` `phi = tan^(-1) (0.6635) = 30.11^@ = 30^@ = (30 xx pi)/(180) = pi/6` `phi = omega t , t = phi/omega = (pi)/(6 xx 2 xx pi xx 69) = 1.21 xx 10^(-3) s = 1.21 ms ` |
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